I am looking for a definition of a meson that does not include the quark model. After some research I have come across this definition:

A meson is a particle that is

(1) believed to be fundamental, not composite,

(2) capable of participating in strong interactions,

(3) a boson.

(source: Semat, H. and Albright, J.R. 1972. Introduction to Atomic & Nuclear Physics. 5th ed. London: Chapman and Hall p.590)

The form of this definition is what I am after, although I think the first point, (1), now no longer applies as mesons are made out of quarks (this book was first published in 1939).

Does anyone know of a similar definition of mesons that does not rely on quarks?

Clarification of what is been asked

Is there a definition of mesons that does not depend on quarks.

  • 2
    $\begingroup$ How about consulting the particle data book for a proper modern collection of information about mesons? Nothing is more stale than a textbook that was already out of date when it was printed. OTOH, it seems to have enough pages to make a good door stop. $\endgroup$
    – CuriousOne
    Commented Jul 14, 2016 at 8:16
  • $\begingroup$ @CuriousOne I have tried but all the once I can find simply give definitions in terms of quarks. $\endgroup$ Commented Jul 14, 2016 at 8:26
  • 3
    $\begingroup$ A meson is a bound state of two (valence) quarks. I can't see how else it could be usefully defined. Can you explain what exactly you're trying to achieve by not referencing quarks? $\endgroup$ Commented Jul 14, 2016 at 8:46
  • $\begingroup$ I would think it would be difficult to find one. I mean one could post a question asking for a definition of an atomic nucleus without referencing protons and neutrons, but one would struggle with that, right? $\endgroup$
    – Matt
    Commented Jul 14, 2016 at 9:08
  • $\begingroup$ @MattS Not necessarily. It'd be easy enough to frame something in terms of the dense blob of mass that sits in the center of an atom. Of course one might argue it's not a particularly useful definition if it doesn't reference protons and neutrons, but for purposes of this question I think we'd have to settle for that. $\endgroup$
    – David Z
    Commented Jul 14, 2016 at 9:10

1 Answer 1


According to the Particle Data Group [1],

Mesons are [strongly interacting particles that] have baryon number $\mathcal{B}=0$. In the quark model, they are $q\bar{q}'$ bound states of quark $q$ and antiquark $\bar{q}'$.

This definition has two sentences, which in fact are somewhat different. The second one defines the quark-model mesons, that you're trying to avoid.

The first one is more interesting. The baryon number $\mathcal{B}$ is, as far as we know, a conserved quantum number. This means that baryons are created only in baryon-antibaryon pairs, while mesons don't have such restriction. This is a reflection of the bosonic statistics for mesons. Therefore, the first sentence is, more or less$\dagger$, equivalent to

A meson is a particle that is

(2) capable of participating in strong interactions,

(3) a boson.

The first definition is more general, in the sense that it includes the so-called exotic mesons:

  • Glueballs, composed only by gluons.
  • Tetraquarks, composed by two quarks and two antiquarks.
  • Mesonic molecules, which are bound states of two quark-model mesons.
  • Hybrid states $q\bar{q}g$ with an excited gluon.
  • Superpositions of several mesonic and exotic states.

In the last years, a number of resonances such as $Z(4430)$, $f_0(980)$ or $f_0(1370)$ are believed to be exotic mesons, but the situation is still unclear.

I think that there is not a consensus regarding whether exotic mesons are to be classified as mesons or as something different. In this sense, your definition of meson might or might not be correct.

$\dagger$ As @fqq points out in the comments below, your definition without (1) allows bosonic nuclei like deuterium to be considered as mesons. The condition $\mathcal{B}=0$ implies bosonic statistics, but not the other way around. Since a nucleus with mass number $A$ has $\mathcal{B}=A$, the way to avoid this complication is just considering $\mathcal{B}=0$ as the definition.

[1] C. Amsler, T. DeGrand, B. Krusche: Quark model in K. A. Olive et al.: 2015 Review of Particle Physics. Chin. Phys. C, 38, 090001 (2014). Link to PDF.

  • $\begingroup$ Isn't the deuteron a strongly interacting boson with $\mathcal{B}\neq 0$? $\endgroup$
    – fqq
    Commented Jul 14, 2016 at 11:49
  • $\begingroup$ @fqq That's why I said more or less. In the OP's definition, deuterons aren't allowed by (1). In the more modern point of view, $\mathcal{B}=0$ implies bosonic statistics, but not the other way around. I'll try to improve the answer later. $\endgroup$
    – Bosoneando
    Commented Jul 14, 2016 at 12:19
  • $\begingroup$ The Particle Data Group's "Note on Scalar Mesons," associated with the $f_0(500)$ in the 2014 Review, is very enlightening. Less decorous people refer to this particle as the "fictitious sigma meson." $\endgroup$
    – rob
    Commented Jul 14, 2016 at 16:05

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